Let $$f(x)$$ be a function satisfying $$f(x) + f(\pi - x) = \pi^2$$, $$\forall x \in \mathbb{R}$$. Then $$\int_0^\pi f(x) \sin x \, dx$$ is equal to

Given $$f(x) + f(\pi - x) = \pi^2$$ for all $$x \in \mathbb{R}$$. We need to find $$\int_0^\pi f(x)\sin x\,dx$$.

Let $$I = \int_0^\pi f(x)\sin x\,dx$$.

Substituting $$x \to \pi - x$$:

$$I = \int_0^\pi f(\pi - x)\sin(\pi - x)\,dx = \int_0^\pi f(\pi - x)\sin x\,dx$$

Adding the two expressions:

$$2I = \int_0^\pi [f(x) + f(\pi - x)]\sin x\,dx = \int_0^\pi \pi^2 \sin x\,dx$$

$$2I = \pi^2[-\cos x]_0^\pi = \pi^2(-(-1) - (-1)) = 2\pi^2$$

$$I = \pi^2$$

The answer is Option C: $$\pi^2$$.